LeetCode 3-Longest Substring Without Repeating Characters

Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

找出字符串中最长的不相同子串

空 长度为0直接返回0

长度为1，返回1

对于大于1的情况

遍历字符串：

定义一个HashSet集合

对字符串中的每个元素，若不存在集合中则，加入集合

若已经存在集合中：假设这个元素是ch，要把加入集合ch之前的元素全部剔除，left加一

while(s.charAt(left)!=ch){
                    set.remove(s.charAt(left));
                    left++;
                }//end while
                left++;

上面就是主要的程序

全部程序

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        HashSet set = new HashSet();
        int left = 0;
        int right = 0;
        int max = 0;
        char ch;
        if(s==null && s.length()==0) return 0;
        if(s.length()==1) return 1;
        for( right=0;right<s.length();right++){
            ch = s.charAt(right);
            if(set.contains(ch)){
                max =Math.max(max,right-left);
                
                while(s.charAt(left)!=ch){
                    set.remove(s.charAt(left));
                    left++;
                }//end while
                left++;
            }else{
                set.add(ch);
                // right
            }//end if
        }//end for
         max = Math.max(max,right-left);
        return max;
    }//end 
}

Python 程序，没看懂

class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        if len(s)==0: return 0
        if len(s)==1: return 1
        lastAppPos={s[0]:0} #last appear position of alphabet 
        longestEndingHere=[1]*len(s) # longest substring ending here 
        for index in xrange(1,len(s)):
            lastPos = lastAppPos.get(s[index],-1)
            if lastPos<index-longestEndingHere[index-1]:
                longestEndingHere[index] = longestEndingHere[index-1] + 1
            else:
                longestEndingHere[index] = index - lastPos
            lastAppPos[s[index]] = index
        return max(longestEndingHere)